1. 228

Summary Ranges

2. 算法

http://www.cnblogs.com/grandyang/p/4603555.html

• O(n)

3. 代码

class  Solution{
public:
    vector<string> summaryRanges(vector<int>& nums){
        vector<string> res;
        int i=0,n=nums.size();
        while(i<n){
            int j=1;
            while(i+j<n && nums[i+j]-nums[i]==j)
                ++j;
            res.push_back(j<=1?to_string(nums[i]):to_string(nums[i])+"->"+to_string(nums[i+j-1]));
            i+=j;
        }
        return res;
    }
};